Question

How do I solve this exponential equation?

3^x^2(9^x)=1/3

Answer 1

can u write it again here in the replies?

Answer 2

wait sory hold on

Answer 3

is it\[\large 3^{x ^{2}}\times9^{x}=\frac{ 1 }{ 3 }\]

Answer 4

yes, thats it

Answer 5

OK, do you know how to solve an equation by taking a natural log of both sides?

Answer 6

no, how do i do that?

Answer 7

Well, I suspect it has been covered in your course because otherwise I'm not sure how we would solve this. You need to understand some things about properties of logarithms. For example:

If a=b then ln(a)=ln(b).

(This is actually true for any base of the log function, but we will use the natural log.)

so if\[\large 3^{x ^{2}}9^{x}=\frac{ 1 }{3 }\]\[\large \ln \left( 3^{x ^{2}}9^{x}\right)=\ln \frac{ 1 }{3 }\]

Answer 8

Now, do you know some properties of logs that will help you simplify that left hand side? You need to start pulling it apart to get those pesky variables OUT OF the exponents, so that you can solve for them.

Answer 9

\[(a^x)' = a^x \ln a\]
if you know how to simplify the statement then you can use this rule to get the derivative like a ninja!

Answer 10

@MrWho .... I don't think this problem has anything to do with the derivative?? Unless I'm missing something, lol.

Answer 11

@MrWho I don't understand that either. @DebbieG would a property be logb(x^n) = n logb^x

Answer 12

Yes, that's what I was getting it! You'll use \[\large \ln x ^{a}=a \ln x\]and also \[\large \ln (MN)=lnM+lnN\]

Answer 13

One you apply those, you should be able to isolate the x and have everything else just constants. :)

Answer 14

@DebbieG thank you!

Answer 15

You're welcome... happy to help :)

Answer 16

@DebbieG after simplifying, I got \[2x \ln 3 + x \ln 9 = \ln 1/3\] but the answer sheet says I'm supposed to get x= -1, -1
so im not sure how to get that answer by further simplifying the ln equation I got

Answer 17

Well, you've re-written the equation but still need to solve it.

However, I just looked at it again and I see an easier way to do this. (I would apologize for making it harder than it needs to be, but hey, it never hurts to practice those math skills, lol. )

So back to what you started with:

\[\large 3^{x ^{2}}9^{x}=\frac{ 1 }{3 }\]

We can re-write this so that every piece of it has a base of 3, right? Rewrite the base 9 term with a base of 3. Then rewrite the 1/3 as a power or 3.

Answer 18

Then remember your rules of exponents, especially\[\large b^mb^n=b^{m+n}\]That will come in handy on the LHS.

Then remember that if:\[\large b^m=b^n\]then m=n.

Answer 19

\[3^{2x }+3^{2x}=3^{-3}\]

Answer 20

so 2x+2x=-3

Answer 21

Not quite...

Answer 22

2x + 2x = -1

Answer 23

because 1/3 = three to the power of negative one?

Answer 24

Look again at that left hand side:\[\large 3^{x ^{2}}9^{x}\]is a product, and re-writing the powers does not turn it into a sum, get me?

So when you convert the 9 to a base of 3, you get:\[\large 3^{x ^{2}}(3^{2})^x=3^{x ^{2}}3^{2x}\]right?

Answer 25

Then using the rule I gave you above, you can write that LHS so that it is just a single exponential form, eg., just one appearance of the base 3 to a power.

Answer 26

\[3^{x ^{2}+2x}\]

Answer 27

OK, good - so now the equation is:\[\large 3^{x2+2x}=3^{-1}\] right?

Now use that property a few posts up, if b^n=b^m then n=m....you'll get yourself a good ol' simple quadratic equation to solve. :)

Answer 28

\[x ^{2}+2x + 1 = (x+1) (x+1)\] so then you set both to zero and get x equals -1 for both sets

Answer 29

There ya go! :)

Answer 30

@DebbieG thank you!!!! :)

Answer 31

You're welcome. :)